Download CHAPTER 7: Potential Energy and Conservation of

Lecture 11: Potential Energy
1
CHAPTER 7: Potential Energy and Conservation of Energy
The most important principle in all of Physics is the Conservation of Energy.
Energy can neither be created or destroyed but only changed from one
form into another.
In Mechanics the two forms of Energy are Kinetic Energy and Potential
Energy. The total Mechanical Energy of a system is equal to the Kinetic
Energy plus the Potential Energy.
Etotal ≡ K + U
The Conservation of Mechanical Energy
In a system where there are no frictional forces acting, the total Mechanical Energy is constant.
When there are no frictional forces acting, we say that there are only conservative
forces acting. Conservative forces include the force of gravity and the spring
(elastic) force.
Kinetic Energy
The Kinetic Energy is always defined the same way for any object. If you
have an object which has a mass m, and that object is moving with a speed v,
then the kinetic energy is always K = 21 mv 2
Potential Energy
The two principle forms of Potential Energy which we deal with in this chapter
are the gravitational potential energy and the elastic potential energy of a spring.
U gravity = mgh (h is the height above some surface)
1
U spring = kx2
2
(x is the compressed or stretched length)
Lecture 11: Potential Energy
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Using the Conservation of Energy to Solve Problems
A mass initially as rest slides down a frictionless inclined plane of height h and
inclination angle Θ. What is the speed of the mass when it reaches the bottom
of the inclined plane ?
Solution by Conservation of Energy
Initial Mechanical Energy
Ki + Ui
=
=
Final Mechanical Energy
Kf + Uf
1
0 + mgh = mvf2 + 0
2
q
=⇒ vf = 2gh (independent of the mass !)
Solution by Equations of Motion and Newton’s Second Law
Calculate the acceleration ax down the plane. Then use the third kinematics
equation (vf2 (x) = v02 + 2ax (x − x0 )).
Newton’s second law gives the acceleration as the force divided by the mass:
Fx
W sin Θ mg sin Θ
=
=
= g sin Θ
m
m
m
The total distance traveled is the length of the plane
ax =
h
sin Θ
Now substitute into the third kinematics equation
x − x0 = x =
vf2 = v02 + 2ax (x − x0 ) = 2g sin Θ(
q
h
) = 2gh
sin Θ
=⇒ vf = 2gh (same answer but more complicated to derive)
Lecture 11: Potential Energy
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Using the Conservation of Energy with Frictional Forces Present
If there are frictional forces present, then the work done against the frictional
(non–conservative) forces is equal to the change (decrease) in the total Mechanical Energy. The total Mechanical Energy is not constant when frictional forces
are present. The Mechanical Energy will decrease because of the work done
against the frictional forces.
Wf riction = (Kf + Uf ) − (Ki + Ui )
Worked Example A 3 kg block slides down a rough incline 1 m in length.
The block starts from rest at the top of the inclined plane, and experiences a
constant force of friction of magnitude 5 N. The angle of the incline is 30o .
Using conservation of energy, determine the speed of the block when it reaches
the bottom of the plane.
Wf riction = (Kf + Uf ) − (Ki + Ui )
1
−f s = ( mvf2 + 0) − (0 + mgh)
2
The force of friction f is given as 5 N, the length s over which it acts is given as 1.0
m, and the initial height of the block h may be found from simple trigonometry
to be 0.5 m.
1
−5 · 1 = 3vf2 − 3 · 9.8 · 0.5
2
vf2 = 6.47 m2 /s2 =⇒ vf = 2.54 m/s
In this example, the total Mechanical Energy was not conserved because of the
non–conservative frictional forces. The decrease in the Mechanical Energy went
into doing work against friction, and that work actually would show up as heat.
Lecture 11: Potential Energy
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Using Conservation of Mechanical Energy in Spring Problems
The principle of conservation of Mechanical Energy can also be applied to systems involving springs. First take a simple case of a mass traveling in a horizontal
direction at constant speed. The mass strikes a spring and the spring begins to
compress slowing down the mass. Eventually the mass stops and the spring is
at its maximum compression. At this point the mass has zero kinetic energy
and the spring has a maximum of potential energy. Of course, the spring will
rebound and the mass will finally be accelerated to the same speed but opposite
in direction. The mass has the same kinetic energy as before, and the spring
returns to zero potential energy.
Spring Potential Energy
If a spring is compressed (or stretched) a distance x from its normal length, then
the spring acquires a potential energy U spring (x):
1
U spring (x) = kx2
2
(k = force constant of the spring)
Worked Example A mass of 0.80 kg is given an initial velocity vi = 1.2 m/s
to the right, and then collides with a spring of force constant k = 50 N/m.
Calculate the maximum compression of the spring.
Solution by Conservation of Energy
Initial Mechanical Energy
=
Final Mechanical Energy
Ki + Ui
=
Kf + Uf
1 2
1
mvi + 0 = 0 + kx2
2
2
=⇒ x = vi
s
v
u
u 0.8
m
= 1.2t
= 0.152 m
k
50
Lecture 11: Potential Energy
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Conservation of Energy and the Motion of a Pendulum
A simple pendulum consists of a sphere of mass m (called a “bob”) attached to
a very light (massless) string of length L. Initially the mass m can be hanging
motionless, straight down from the string. Then the mass m is displaced upward
such that the string makes an angle θ0 with the vertical direction. The mass is
then released. The mass goes back to the vertical position and acquires kinetic
energy. Because the mass is moving it does not simply stop when the string is
at the vertical position, but rather continues to the other side until it reaches
the same height at which it was first released. It then repeats the motion in the
opposite direction.
In the limit that there is no friction, the mass of a pendulum will constantly
swing back and forth. What is really happening is that there is a continuous
transformation of potential energy into kinetic energy, and then kinetic energy
back into potential energy. As long as none of the mechanical energy is lost to
friction, the motion should continue forever. The motion of the pendulum goes
by the name “oscillation” or “simple harmonic motion”, and will be studied in
greater detail in Chapter 12. For the moment, we can calculate the maximum
speed of the mass. The maximum speed of the mass occurs when the mass
is at the lowest point of the motion. The zero of potential energy is defined
conveniently at the top of the string.
Initial Mechanical Energy
=
Ka + Ua
=
Final Mechanical Energy
Kb + Ub
1
0 − mgL cos θ0 = mvb2 − mgL
2
q
vb = 2gL(1 − cos θ0 ) (independent of the mass)
Lecture 11: Potential Energy
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Conservation of Energy and the Loop–the–Loop
The loop–the–loop consists of a curved track whose initial point is located a
distance h above ground level. The curve goes into a vertical circle of radius R
with its bottom most point at ground level. The track and the vertical circle
are assumed to be frictionless. Describe qualitatively the motion of the particle
released from rest at the top of the track in terms of the potential and kinetic
energies, and the forces acting on the particle. Why does the particle sometimes leave the vertical circle before reaching the top most point ? Obtain an
expression for the minimum value of h such that the particle will not fall off the
vertical circle portion of the track.
Since there are no frictional forces acting, the total Mechanical Energy is conserved. The particle was released from rest at the top, so the total energy is
given by the initial potential energy:
Etotal = mgh
At any lower point y the particle will have some speed v(y) such that the total
energy is conserved
Etotal = Ki + Ui = Ky + Uy
q
1 2
0 + mgh = mv (y) + mgy =⇒ v(y) = 2g(h − y)
2
In particular, at the top of the vertical circle y = 2R, the speed is given by
q
v(y = 2R) = 2g(h − 2R)
But if v is too small (because h is too close to 2R) the particle will not be able
to stay on the vertical circle once it gets past the midway point of the circle.
The only way it can stay on the track is for the centripetal force to be at least
as great as the weight force at all points along the top half of the circle.
Lecture 11: Potential Energy
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Conservation of Energy and the Loop–the–Loop
Now at the top of the circle, the centripetal force must be at least as great as
the weight force on the particle. If not, then the particle will have a net vertical
acceleration. This sets a minimum value on the speed of the particle at the top
of the circle
q
v(y) = 2g(h − y) (from energy conservation)
At the top of the vertical circle y = 2R, the speed is given by
q
v(y = 2R) = 2g(h − 2R) (from energy conservation)
But if v is too small (because h is too close to 2R) the particle will not be able
to stay on the vertical circle once it gets past the midway point of the circle.
The only way it can stay on the track is for the centripetal force to be at least
as great as the weight force at all points along the top half of the circle.
Fcentripetal
mv 2
=
= mg
R
q
=⇒ vminimum = gR (from centripetal force)
One can then find the minimum value of h which will allow this minimum value
of v
q
q
vminimum = gR = 2g(hminimum − 2R)
5
=⇒ hminimum = R
2
If you can understand this problem, you know a lot.